Question: Simplify the following expression: $y = \dfrac{2x^2- 7x+3}{2x - 1}$
Answer: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(2)}{(3)} &=& 6 \\ {a} + {b} &=& &=& {-7} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $6$ and add them together. The factors that add up to ${-7}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-1}$ and ${b}$ is ${-6}$ $ \begin{eqnarray} {ab} &=& ({-1})({-6}) &=& 6 \\ {a} + {b} &=& {-1} + {-6} &=& -7 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({2}x^2 {-1}x) + ({-6}x +{3}) $ Factor out the common factors: $ x(2x - 1) - 3(2x - 1)$ Now factor out $(2x - 1)$ $ (2x - 1)(x - 3)$ The original expression can therefore be written: $ \dfrac{(2x - 1)(x - 3)}{2x - 1}$ We are dividing by $2x - 1$ , so $2x - 1 \neq 0$ Therefore, $x \neq \frac{1}{2}$ This leaves us with $x - 3; x \neq \frac{1}{2}$.